Recall that Casorati-Weierstrass theorem says that given $f$ is holomorphic in the punctured disc $D_r(z_0) - \{z_0\}$ and has an essential singularity at $z_0$. Then, the image of $D_r(z_0) - \{z_0\}$ under $f$ is dense in the complex plane.

When does the converse fail, if ever? It seems like IF the image of $f$ around a neighborhood of $z_0$ is dense in the complex plane, then neither the limit of $f$ will exist as $z \rightarrow z_0$, nor will it be unbounded because then it would be impossible for me to exhibit convergent sequences as $z\rightarrow z_0$. So what am I missing/misunderstanding?

  • 2
    $\begingroup$ I'd say the converse is the trivial direction. If the function has a removable singularity at a point, it's obviously bounded near it, so the image can't be dense. Similarly, if the function has a pole then its modulus must be bounded away from zero and the image will now miss the nbhd of 0. I suspect this is why the theorem is not stated as an equivalence. $\endgroup$
    – Marek
    Commented Jan 1, 2014 at 11:30
  • $\begingroup$ @Marek Thank you for the remark. $\endgroup$
    – compy
    Commented Jan 1, 2014 at 11:43
  • $\begingroup$ Careful. If you're looking at only one $r > 0$, it can easily be that $f(D_r(z_0)\setminus\{z_0\})$ is dense in the plane but $z_0$ is a removable singularity or a pole. You need that $f(D_r(z_0)\setminus\{z_0\})$ is dense for all $r > 0$ (small enough that $f$ is defined on the punctured disk). $\endgroup$ Commented Jan 1, 2014 at 12:50
  • $\begingroup$ @DanielFischer, thanks. Were you thinking of, say for $z_0 = 0$ and $r_0$ real, $f(re^{i\theta}) = \exp(\frac{1}{r-r_0}e^{i\theta})$? $\endgroup$
    – compy
    Commented Jan 1, 2014 at 16:18
  • $\begingroup$ No. That doesn't look holomorphic at first glance, by the way. I was thinking of something like $f = \lambda \circ T$, where $\lambda$ is the modular function on the upper half plane, and $T$ is a Möbius transformation mapping the disk onto the upper half plane. Then $f(D_r(z_0)\setminus\{z_0\}) = \mathbb{C}\setminus \{0,1\}$, and you only get a non-dense image when you restrict to a radius $\rho < r$. $\endgroup$ Commented Jan 1, 2014 at 16:26

1 Answer 1


You're not missing anything; the converse is always true.

EDIT: In fact there is a technicality here; we have to assume that $f$ is defined on the closed disc, or the behavior of $f$ on the boundary comes into play.

  • $\begingroup$ As stated, the Converse doesn't seem to be true. OP is assuming that the image of some deleted neighbourhood is dense, according to @Daniel Fischer in the comments this isn't true. I have not been able to find a proof of this either after working for a while. $\endgroup$ Commented Feb 16 at 13:16
  • $\begingroup$ @AnshumanAgrawal It's fine if we work with closed discs, but since the disc is implicitly assumed open I added a note. $\endgroup$ Commented Feb 16 at 13:55
  • $\begingroup$ what happens in the case of open disc? I've been trying to resolve this, I thought I had a proof but it doesn't really work, I started to think the result wasn't true in that form and this seems to be supported by a comment above. I would appreciate a resolution if you know it. $\endgroup$ Commented Feb 19 at 16:27
  • $\begingroup$ @AnshumanAgrawal I'm under the impression that Daniel Fischer's construction above gives a counterexample. $\endgroup$ Commented Feb 20 at 10:55

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